Centripetal Force

Proposition 41 Problem 28: Supposing a centripetal force of any kind, and granting the quadratures of curvilinear figures, it is required to find as well the trajectories in which bodies will move, as the times of their motions in the trajectories found.

Let any centripetal force tend to the centre C, and let it be required to find the trajectory VIKk. Let there be given the circle VR, described from the centre C with any interval CV.

From the same centre describe any other circles ID, KE cutting the trajectory in I and K, and the right line CV in D and E. Then draw the right line CNIX cutting the circles KE, VR in N and X, and the right line CKY meeting the circle VR in Y.

Let the points I and K be indefinitely near; and let the body go on from V through I and K to k; and let the point A be the place from whence another body is to fall, so as in the place D to acquire a velocity equal to the velocity of the first body in I. And things remaining as in Prop. XXXIX, the lineola IK, described in the least given time will be as the velocity, and therefore as the right line whose square is equal to the area ABFD, and the triangle ICK proportional to the time will be given, and therefore KN will be reciprocally as the altitude IC; that is (if there be given any quantity Q, and the altitude IC be called A), as …

This quantity … call Z.

A Suppose the magnitude of Q to be such that in some case .. may be to Z as IK to KN, and then in all cases … will be to Z as IK to KN, and ABFD to ZZ as IK² to KN², and by division ABFD - ZZ to ZZ as IN² to KN², and therefore .. to Z; or … as IN to KN; and therefore A x KN will be equal to …

Therefore since YX × XC is to A × KN as CX², to AA, the rectangle XY × XC will be equal to …

Therefore in the perpendicular DF let there be taken continually Db, Dc equal to

respectively, and let the curve lines ab, ac, the foci of the points b and c, be described: and from the point V let the perpendicular Va be erected to the line AC, cutting off the curvilinear areas VDba, VDca, and let the ordinates Ez, Ex, be erected also.

Then because the rectangle Db × IN or DbzE is equal to half the rectangle A × KN, or to the triangle ICK; and the rectangle Dc × IN or DcxE is equal to half the rectangle YX × XC, or to the triangle XCY; that is, because the nascent particles DbzE, ICK of the areas VDba, VIC are always equal.

The nascent particles DcxE, XCY of the areas VDca, VCX are always equal: therefore the generated area VDba will be equal to the generated area VIC, and therefore proportional to the time.

The generated area VDca is equal to the generated sector VCX. If, therefore, any time be given during which the body has been moving from V, there will be also given the area proportional to it VDba.

Thence, will be given the altitude of the body CD or CI; and the area VDca, and the sector VCX equal thereto, together with its angle VCI. But the angle VCI, and the altitude CI being given, there is also given the place I, in which the body will be found at the end of that time. Q.E.I.

Corolllary 1

Hence the greatest and least altitudes of the bodies, that is, the apsides of the trajectories, may be found very readily. For the apsides are those points in which a right line IC drawn through the centre falls perpendicularly upon the trajectory VIK; which comes to pass when the right lines IK and NK become equal; that is, when the area ABFD is equal to ZZ.

Corolllary 2

So also the angle KIN, in which the trajectory at any place cuts the line IC, may be readily found by the given altitude IC of the body: to wit, by making the sine of that angle to radius as KN to IK that is, as Z to the square root of the area ABFD.

Corolllary 3

If to the centre C, and the principal vertex V, there be described a conic section VRS; and from any point thereof, as R, there be drawn the tangent RT meeting the axis CV indefinitely produced in the point T.

Then joining CR there be drawn the right line CP, equal to the abscissa CT, making an angle VCP proportional to the sector VCR; and if a centripetal force, reciprocally proportional to the cubes of the distances of the places from the centre, tends to the centre C; and from the place V there sets out a body with a just velocity in the direction of a line perpendicular to the right line CV; that body will proceed in a trajectory VPQ, which the point P will always touch.

Therefore if the conic section VRS be an hyberbola, the body will descend to the centre; but if it be an ellipsis, it will ascend perpetually, and go farther and farther off in infinitum.

On the contrary, if a body endued with any velocity goes off from the place V, and according as it begins either to descend obliquely to the centre, or ascends obliquely from it, the figure VRS be either an hyperbola or an ellipsis, the trajectory may be found by increasing or diminishing the angle VCP in a given ratio.

The centripetal force becoming centrifugal, the body will ascend obliquely in the trajectory VPQ, which is found by taking the angle VCP proportional to the elliptic sector VRC, and the length CP equal to the length CT, as before.

All these things follow from the foregoing Proposition, by the quadrature of a certain curve, the invention of which, as being easy enough, for brevity’s sake I omit.